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A problem set focused on the hardy-weinberg equilibrium (hwe) for various genetic traits. It includes calculations of genotype and allele frequencies, as well as predictions based on the hwe law. The problem set covers topics such as thalassemia, pku, and blood types.
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Hardy-Castle-Weinberg (20 pts total)
Genotype AA AS SS Total Number 189 89 9 287 Characterize this population by its genotype frequencies. Characterize the gene pool by the allele frequenciesfor A and S. Using Hardy-Weinberg Law, predict the genotype frequencies based on the allele frequencies. ( pts) Observed Genotype Frequencies: Freq(AA) = 189/287 = 0.66Freq(AS) = 89/287 = 0. Freq(SS) = 9/287 = 0. Allele Frequencies: Freq(A) =Freq(S) = pq = 189/287 + ½(89/287) = 0.81= 9/287 + ½(89/287) = 0. (Or use p + q = 1 to solve for q ) Expected HWE genotype frequencies p 22 pq^ = (0.81) = 2(0.81)(0.19) = 0.3 2 = 0. q^2 = (0.19) 2 = 0.
Population 1:Population 2: AAAA 162;18; Aa Aa 84; 36; aa aa 98 2 Now suppose these two populations are combined to form a new population: Population 3: AA 180; Aa 120; aa 100 a. Can the differences in genotype frequencies in populations 1 and 2 be explained by differences in the mating systems with respect to this gene locus? (2 pts) Pop 1 Genotype frequencies AA = 162/200 = 0.81; Aa = 36/200 = 0.18; aa = 2/200 = 0.01Allele frequencies p Expected H-W frequencies A^ = 0.81 +1/2(0.18) = 0.9; p 2 = (0.9) 2 = 0.81; 2^ qpq^ a = 0.01 + 1/2(0.18) = 0.1 = 2(0.9)(0.1) = 0.18; q 2 = (0.1)^2 = 0. This population is in HW equilibrium Pop 2 Genotype frequencies AA = 18/200 = 0.09; Aa = 84/200 = 0.42; aa = 98/200 = 0. Allele frequenciesExpected H-W frequencies p (^) A = 0.09 +1/2(0.42) = 0.03; p 2 = (0.3) 2 = 0.09; 2 qpq (^) a = 0.49 + 1/2(0.42) = 0.7 = 2(0.3)(0.7) = 0.42; q 2 = (0.7) (^2) = 0. This population is in HW equilibrium Since both populations are in HW expected proportions they do not violate the assumption of randommating. There is no evidence that differences in the mating system are responsible for the differences in genotype frequencies. b. Is population 3 in H-W equilibrium? If not, indicate whether heterozygotes are in excess or indeficiency. (2 pts) Pop 3 Genotype frequencies AA = 180/400 = 0.45; Aa = 120/400 = 0.3; aa = 100/400 = 0. Allele frequencies p (^) A = 0.45 +1/2(0.3) = 0.6; q (^) a = 0.25 + 1/2(0.3) = 0. Expected H-W frequencies p 2 = (0.6) 2 = 0.36; 2 pq = 2(0.6)(0.4) = 0.0.48; q 2 = (0.4) 2 = 0. The HW expected genotype frequencies do not match the observed genotype frequencies so thispopulation is not in HW equilibrium. There is an excess of homozygotes and a deficit of heterozygotes. c. For how many generations will the effect of this single episode of admixture of populations bedetectable in a population established from population 3 with respect to genotype frequencies if mating is random with respect to this locus? (2 pts) After one generation of random mating this populations will be in HW proportions and will stay inthose proportions. You can only see the mixture in the first mixed generation.